First slide

Simple hormonic motion

Question

A particle is executing SHM with time period π sec along a straight line. At t=0, it has started its motion from one of its extreme position. Then the minimum time after which its kinetic energy will be equal to its potential energy is

Moderate

Solution

Equation of SHM is $x = Acosωt$
At time t , potential energy, $U = \frac{1}{2} {mω}^{2} x^{2} = \frac{1}{2} {mω}^{2} A^{2} \cos^{2} ωt$
And kinetic energy, $K = \frac{1}{2} {mω}^{2} (A^{2} - x^{2}) = \frac{1}{2} {mω}^{2} A^{2} \sin^{2} ωt$
$When K = U, \frac{1}{2} {mω}^{2} A^{2} \sin^{2} ωt = \frac{1}{2} {mω}^{2} A^{2} \cos^{2} ωt \Rightarrow tanωt = \pm 1$
$\begin{array}{l} \Rightarrow ωt = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}, . ....... etc \\ ∴ t_{\min} = \frac{π}{4 ω} = \frac{π}{4 \times \frac{2 π}{π}} = \frac{π}{8} \sec \end{array}$

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